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Combinatorics question.

  1. Sep 27, 2006 #1
    You're playing standard bridge with three other people. If you know that you and your partner have 10 hearts altogether, then what is the probability that the remaining 3 hearts are all in the same hand ?

    Is it comb(3,3)*comb(23,10)/comb(26,13) ? Since there is only 1 way of selecting 3 hearts and com(23,10) ways of selecting the other 10 cards and comb(26,13) different combinations for the opponents cards ?
  2. jcsd
  3. Sep 27, 2006 #2


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    So in bridge, is every card in somebody's hand? And each person always has 13 cards?

    What you've said is the probability that the remaining three hearts in a particular person's hand. But there are two possible hands it could be in. What you're looking for is:
    1. The number of ways for person A to select 3 hearts and 10 other cards (you basically have this already)
    2. The number of ways for person A not to select 3 hearts, and then for person B to select the remaining cards.
    Add up 1 and 2 and divide by the total number of ways for A and B to select their cards.
  4. Sep 27, 2006 #3
    Thanks that is what I thought. he he he :)
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